Chapter 9.4, Problem 63E

a.

To determine

To write a set of parametric equations that model the path of the baseball.

Answer to Problem 63E

  $x=\left(\frac{440}{3}\cos \theta \right)t$ and $y=3+\left(\frac{440}{3}\sin \theta \right)t-16t^2$ .

Explanation of Solution

Given: The ball is hit at a point 3 ft above the ground at an angle $\theta$ with the horizontal at a speed of 100 miles per hour.

Calculation:

The velocity of the ball in feet per second can be given as $100\text{miles/hour }=100\times \frac{5280}{60\cdot 60}=\frac{440}{3}\text{feet/sec}$ .

Therefore, the parametric equations for the path of the ball can be given as:

  $x=\left(v_0\cos \theta \right)t=\left(\frac{440}{3}\cos \theta \right)t$ ,

  $y=h+\left(v_0\sin \theta \right)t-16t^2=3+\left(\frac{440}{3}\sin \theta \right)t-16t^2$ .

That is,

  $x=\left(\frac{440}{3}\cos \theta \right)t$ and $y=3+\left(\frac{440}{3}\sin \theta \right)t-16t^2$ .

b.

To determine

To graph the path of the ball when $\theta =15°$ by using a graphing utility.

Explanation of Solution

Given:

  $\theta =15°$

Graph:

The parametric equation for $\theta =15°$ becomes:

  $x=\left(\frac{440}{3}\cos 15°\right)t$ and $y=3+\left(\frac{440}{3}\sin 15°\right)t-16t^2$ .

Now, graph these parametric equations using graphing utility as shown below.

  

From the graph it can be seen that the ball hits the ground at a distance of around 350 ft. So, it is not a home run.

c.

To determine

To graph the path of the ball when $\theta =23°$ by using a graphing utility.

Explanation of Solution

Given:

  $\theta =23°$

Graph:

The parametric equation for $\theta =23°$ becomes:

  $x=\left(\frac{440}{3}\cos 23°\right)t$ and $y=3+\left(\frac{440}{3}\sin 23°\right)t-16t^2$ .

Now, graph these parametric equations using graphing utility as shown below.

  

From the graph it can be seen that the ball hits the ground at a distance of around 490 ft. The height of the ball at a distance of 408 ft is around 13 ft. So, it is a home run.

d.

To determine

To find the minimum value of the angle $\theta$ for the hit to become a home run.

Answer to Problem 63E

  $\theta =19.31°$

Explanation of Solution

Given: The parametric equations $x=\left(\frac{440}{3}\cos \theta \right)t$ and $y=3+\left(\frac{440}{3}\sin \theta \right)t-16t^2$ .

Calculation:

The home will be hit if $x=408$ and $y\ge 7$ .

Now,

  $x=408\Rightarrow \left(\frac{440}{3}\cos \theta \right)t=408\Rightarrow t=\frac{408}{\left(\frac{440}{3}\cos \theta \right)}$ .

And,

  $\begin{array}{c}y\ge 7\\ 3+\left(\frac{440}{3}\sin \theta \right)\left(\frac{408}{\left(\frac{440}{3}\cos \theta \right)}\right)-16{\left(\frac{408}{\left(\frac{440}{3}\cos \theta \right)}\right)}^2\ge 7\\ 3+408\tan \theta -\frac{374544}{3025}{\sec }^2\theta \ge 7\\ 408\tan \theta -\frac{374544}{3025}{\sec }^2\theta \ge 4\end{array}$

Now, find the solutions of this inequality, graph the equations $y=408\tan x-\frac{374544}{3025}{\sec }^{2}x$ and $y=4$ , and observe for which value of x, the equation $y=408\tan x-\frac{374544}{3025}{\sec }^{2}x$ is greater than or equal to $y=4$ .

  

From the graph it can observed that the equation $y=408\tan x-\frac{374544}{3025}{\sec }^{2}x$ has same value as that of $y=4$ at $x=0.337$ .

So, the minimum angle required to hit a home run is $\theta =0.337\text{ rad}$ or $\theta =19.31°$ .